### Related Formula p = fracUc$p = \frac{U}{c}$ ### Core Logic For an electromagnetic wave incident on a surface that is completely absorbed, the total momentum transferred is equal to the total energy transferred divided by the speed of light in vacuum (c$c$). ### Step 1: Calculate Momentum Given total energy E = 6.48 times 10^5 mathrm~J$E = 6.48 \times 10^5 \mathrm{~J}$. Speed of light c = 3 times 10^8 mathrm~m/s$c = 3 \times 10^8 \mathrm{~m/s}$. p = fracEc = frac6.48 times 10^53 times 10^8$p = \frac{E}{c} = \frac{6.48 \times 10^5}{3 \times 10^8}$ p = 2.16 times 10^-3 mathrm~kg~m/s$p = 2.16 \times 10^{-3} \mathrm{~kg~m/s}$ ### Pattern Recognition Always check for "complete absorption" versus "perfect reflection". For complete absorption, p = E/c$p = E/c$. For perfect reflection, p = 2E/c$p = 2E/c$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Core Logic Match each law with its corresponding mathematical expression (Maxwell's equations). (A) Gauss's law of magnetostatics: The net magnetic flux through any closed surface is zero. oint vecB cdot mathrmdveca = 0$\oint \vec{B} \cdot \mathrm{d}\vec{a} = 0$ (Matches II). (B) Faraday's law of electromagnetic induction: The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux. oint vecE cdot mathrmdvecl = -fracmathrmdmathrmdt int vecB cdot mathrmdveca$\oint \vec{E} \cdot \mathrm{d}\vec{l} = -\frac{\mathrm{d}}{\mathrm{d}t} \int \vec{B} \cdot \mathrm{d}\vec{a}$ (Matches III). (C) Ampere's law: The line integral of the magnetic field around a closed loop is proportional to the electric current passing through the loop. oint vecB cdot mathrmdvecl = mu_0 I$\oint \vec{B} \cdot \mathrm{d}\vec{l} = \mu_0 I$ (Matches IV). (D) Gauss's law of electrostatics: The electric flux through any closed surface is proportional to the enclosed electric charge. oint vecE cdot mathrmdveca = frac1varepsilon_0 int rho mathrmdV$\oint \vec{E} \cdot \mathrm{d}\vec{a} = \frac{1}{\varepsilon_0} \int \rho \mathrm{d}V$ (Matches I). ### Step 1: Final Match A rightarrow$\rightarrow$ II B rightarrow$\rightarrow$ III C rightarrow$\rightarrow$ IV D rightarrow$\rightarrow$ I This matches option (4). ### Pattern Recognition These are the fundamental Maxwell equations in integral form. Memorizing their direct mappings guarantees quick marks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves Class 12 Physics: Electromagnetic Induction

### Related Formula B_0 = fracE_0C$B_0 = \frac{E_0}{C}$ hatC = hatE times hatB$\hat{C} = \hat{E} \times \hat{B}$ ### Core Logic In an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by E_0 = c B_0$E_0 = c B_0$. Thus, B_0 = E_0 / C$B_0 = E_0 / C$. The direction of wave propagation is given by the cross product of the electric field and magnetic field vectors: hatC = hatE times hatB$\hat{C} = \hat{E} \times \hat{B}$. ### Step 1: Determine Wave Direction and Magnetic Field Direction Given the phase term (omega t - kz)$(\omega t - kz)$, the wave propagates in the +z$+z$ direction, so hatC = hatk$\hat{C} = \hat{k}$. The electric field oscillates in the +x$+x$ direction, so hatE = hati$\hat{E} = \hat{i}$. We know: hatk = hati times hatB$\hat{k} = \hat{i} \times \hat{B}$ Since hati times hatj = hatk$\hat{i} \times \hat{j} = \hat{k}$, the magnetic field must oscillate in the +y$+y$ direction (hatj$\hat{j}$). ### Step 2: Construct Final Vector The full magnetic field vector shares the same phase and applies the above amplitude and direction: vecB = fracE_0C cos(omega t - kz) hatj$\vec{B} = \frac{E_0}{C} \cos(\omega t - kz) \hat{j}$ ### Pattern Recognition Phase remains identical. Amplitude scales by 1/c$1/c$. Direction satisfies the right-hand triad (vecE, vecB, vecv)$(\vec{E}, \vec{B}, \vec{v})$ where vecv = vecE times vecB$\vec{v} = \vec{E} \times \vec{B}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves

### Related Formula rho_E = frac12varepsilon_0 E_rms^2$\rho_E = \frac{1}{2}\varepsilon_0 E_{rms}^2$ rho_B = fracB_rms^22mu_0$\rho_B = \frac{B_{rms}^2}{2\mu_0}$ p_r = fracIc$p_r = \frac{I}{c}$ (radiation pressure for completely absorbing surface) ### Core Logic Statement I: In an EM wave, energy is stored in both the electric and magnetic fields. The energy densities are equal: U_E = U_B$U_E = U_B$ because E = cB$E = cB$ and c^2 = frac1mu_0 epsilon_0$c^2 = \frac{1}{\mu_0 \epsilon_0}$. This is a correct fact. Statement II: Electromagnetic waves carry momentum (p = U/c$p = U/c$). When they strike a surface, they transfer this momentum, creating radiation pressure. This is also a correct fact. ### Step 1: Final Evaluation Since both facts are universally true theoretical properties of EM waves, both statements are correct. ### Pattern Recognition Standard NCERT theory. Energy density is strictly symmetric 50-50 between E and B. Momentum transfer to$\to$ pressure is the definitive signature of the particle-like nature (photons) of EM waves. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Waves